A fly wheel rotating about a fixed axis has a kinetic energy of `360 J`. When its angular speed is `30 rad s^(-1)`. The moment of inertia of the wheel about the axis of rotation is

Options

A. `0.6kgm^(-2)`
B. `0.15kgm^(-2)`
C. `0.8kgm^(-2)`
D. `0.75kgm^(-2)`

Explanation

To find the moment of inertia of the flywheel, we can use the formula for kinetic energy of a rotating object:

\[ KE = \frac{1}{2} I \omega^2 \]

Where:
- \( KE \) is the kinetic energy,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular speed.

Given:
- \( KE = 360 \, \text{J} \)
- \( \omega = 30 \, \text{rad/s} \)

We can rearrange the formula to solve for \( I \):

\[ I = \frac{2 \cdot KE}{\omega^2} \]

Now, substituting the given values into the equation:

1. Calculate \( \omega^2 \):
\[ \omega^2 = (30 \, \text{rad/s})^2 = 900 \, \text{rad}^2/\text{s}^2 \]

2. Substitute \( KE \) and \( \omega^2 \) into the equation for \( I \):
\[ I = \frac{2 \cdot 360 \, \text{J}}{900 \, \text{rad}^2/\text{s}^2} \]

3. Calculate the numerator:
\[ 2 \cdot 360 = 720 \]

4. Now, divide by \( 900 \):
\[ I = \frac{720}{900} = 0.8 \, \text{kg m}^2 \]

Thus, the moment of inertia of the wheel about the axis of rotation is:

\[ I = 0.8 \, \text{kg m}^2 \]